class 的重载成员函数能否依赖于 class 的重载构造函数的结果？

Question

我有一个带有重载构造函数的 class，其中构造函数的每个版本都为 class 初始化一组不同的私有属性。我还有一个 class 的 public 成员函数，它将根据 class 的私有属性执行一些操作。我想重载成员函数，以便当我从主函数调用它时，它将执行一个操作和 return 一个值。根据相应构造函数的确切结果，每个操作都会有所不同。这可能吗？我如何在 C++ 中实现它？这是一些试图表达想法的错误代码：

class someClass {
    
    double var1, var2, var3, var4, var5;

    public:
    someClass(double in1) {
        // operations that initialize var1
    }

    someClass(double in1, double in2) {
        // operations that initialize var1 and var2
    }

    someClass(double in1, double in2, double in3) {
        // operations that initialize var1, var2 and var3
    }

    someClass(double in1, double in2, double in3, double in4) {
        // operations that initialize var1, var2, var3 and var4
    }

    someClass(double in1, double in2, double in3, double in4, double in5) {
        // operations that initialize var1, var2, var3, var4 and var5
    }


    double calcVal() {
        return in1 + in3;
        // this one is executed if the 1st constructor was called
    }

    double calcVal() {
        return in1 + in2;
        // this one is executed if the 2nd constructor was called
    }

    double calcVal() {
        return in1 + in2 + in3;
        // this one is executed if the 3rd constructor was called
    }

    double calcVal() {
        return in1 + in2 + in3 + in4;
        // this one is executed if the 4th constructor was called
    }

    double calcVal() {
        return in1 + in2 + in3 + in4 + in5;
        // this one is executed if the 5th constructor was called
    }
}

Answer 1

我不确定我是否得到你想要的，但根据你的描述，你可以根据调用的构造函数简单地分配一些枚举。然后在calcVal()成员函数中测试：

class someClass {
    enum class constr { cons1, cons2, cons3, cons4, cons5 };
    double var1, var2, var3, var4, var5;
    constr c;
 
    public:
    someClass(double in1) {
       c = constr::cons1; // and initialisation, of course...
    }

    someClass(double in1, double in2) {
        c = constr::cons2; // more here
    }

    someClass(double in1, double in2, double in3) {
        c = constr::cons3; // more here
    }

    someClass(double in1, double in2, double in3, double in4) {
        c = constr::cons4; // more here
    }

    someClass(double in1, double in2, double in3, double in4, double in5) {
        c = constr::cons5; // more here
    }


    double calcVal() {
        switch(c)
        { 
           case constr::cons1:  return var1 + var3;
           case constr::cons2:  return var1 + var2;
        // you get the idea...
        }
    }
}

Answer 2

对我来说，这看起来像是带有虚函数的继承。

struct someClass {
     virtual ~someClass() {}
     virtual double calcVal() = 0;
};

struct classWithVar1 : someClass {
    double var1;
    classWithVar1(double in1) : var1(in1) {}
    double calcVal() override { return var1; }
};

struct classWithVar2 : someClass {
    double var1, var2;
    classWithVar2(double in1, double in2) : var1(in1), var2(in2) {}
    double calcVal() override { return var1 + var2; }
};

/* etc. */

class 的重载成员函数能否依赖于 class 的重载构造函数的结果？

Can an overloaded member function of a class depend on the outcome of an overloaded constructor of that class?

c++

overloading

constructor-overloading